## Precalculus (6th Edition) Blitzer

Solutions of the equation are $\frac{\pi }{3}$ and $\frac{5\pi }{3}$.
On solving the equation, \begin{align} & \sin x+2\sin \left( \frac{x}{2} \right)=\cos \left( \frac{x}{2} \right)+1 \\ & 2\sin \left( \frac{x}{2} \right)\cdot \cos \left( \frac{x}{2} \right)+2\sin \left( \frac{x}{2} \right)=\cos \left( \frac{x}{2} \right)+1 \\ & 2\sin \left( \frac{x}{2} \right)\left( \cos \left( \frac{x}{2} \right)+1 \right)-\left( \cos \left( \frac{x}{2} \right)+1 \right)=0 \\ & \left( \cos \left( \frac{x}{2} \right)+1 \right)\left( 2\sin \left( \frac{x}{2} \right)-1 \right)=0 \end{align} Simplified, \begin{align} & \cos \left( \frac{x}{2} \right)=-1 \\ & \cos \left( \frac{x}{2} \right)=\cos \left( \frac{3\pi }{2} \right) \\ & x=3\pi \end{align} But this does not lie in the interval $[0,2\pi )$. So, it will not be considered.   Further, \begin{align} & \sin \left( \frac{x}{2} \right)=-\frac{1}{2} \\ & \sin \left( \frac{x}{2} \right)=\sin \left( \frac{\pi }{6} \right) \end{align} Or $\sin \left( \frac{x}{2} \right)=\sin \left( \frac{5\pi }{6} \right)$ This implies, $x=\frac{\pi }{3},\frac{5\pi }{3}$ Thus, $\frac{\pi }{3}$ and $\frac{5\pi }{3}$ only lie in the interval $[0,2\pi )$.